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I recently started reading functional analysis on my own and have come about dual spaces and cannot get an intuitive understanding about them. This is where my intuition breaks down while understanding duals of $L^p[0,1]$ spaces for $1\le p<\infty$:

Dual of $L^p[0,1]$ is $L^q[0,1]$ where $1/p+1/q=1$. Here, a dual space consists of 'Linear functionals' (say $l(.)$) that map elements in $L^p[0,1]$ to $R$, whereas $L^q[0,1]$ contains functions (say $f(.)$) that map real numbers to real numbers. In otherwords, the map $l(.)$ works on functions and the map $f(.)$ works on real numbers. How can the linear functionals (dual space) be the same as an elements in $L^q[0,1]$.

I do not have a rigorous mathematics background, If this question is vague or trivial please provide references of better sources which contain simple examples.

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    $\begingroup$ They are not the same, just a 1 to 1 correspondence. $\endgroup$ – Shuhao Cao May 5 '13 at 21:40
  • $\begingroup$ A dual space does not consists of linear functionals, but it consists of all continuous linear functionals. The dual space is a subspace (proper in case of infinite dimensional vector spaces) of the space of all linear functionals. $\endgroup$ – Damian Sobota May 5 '13 at 22:03
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You're absolutely correct in your reasoning. From a set-theoretical perspective, the dual space of $L^p$ and the space $L^q$ with $\frac{1}{p} + \frac{1}{q} = 1$ are indeed not the same.

However, there's a one-to-one correspondence between those two sets, and this correspondence is compatible with all the usual operations you'd perform on elements of a vector space. Such a correspondence is called a isomorphism. In the case of $L^p$ and $L^q$, you can define a function $\lambda_g$ for every $g \in L^q$ which maps elements of $L^p$ to $\mathbb{R}$. $$ \lambda_g \,:\, L^p \to \mathbb{R} \,,\, f \to \int_\mathbb{R} f(x)g(x) d\mu $$

It turns out that $\lambda_g$ is a continous map from $L^p$ to $\mathbb{R}$ for every $g \in L^q$, i.e. $\lambda_g$ is an element of the dual space of $L^p$ for every $g \in L^q$. And furthermore, for every elment $\lambda$ in the dual space of $L^p$, there's a $g \in L^q$ such that $\lambda$ and $\lambda_g$ behave identically on $L^p$, meaning for every $f \in L^p$, $\lambda(f) = \lambda_g(f)$.

For all pratical purposes, you may thus call $L^q$ the dual space of $L^p$, if you remember the definition $\lambda_g$ which tells you how to convert functions from $L^p$ to $\mathbb{R}$ into elements of $L^q$ and back.

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  • $\begingroup$ The identification of the dual of $L_p$ with $L_q$ is also an isometry. This should not be overlooked. $\endgroup$ – David Mitra May 5 '13 at 23:07
  • $\begingroup$ Thank you very much. Now I see the correspondence. $\endgroup$ – nadurthi May 5 '13 at 23:20
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this is a representation theorem. this means that each continuous functional is represented by an element in $L^q$ and viceversa.

EDIT after the comment by Damian Sobota:

the fact that every countinuos functional is represented by an element in $L^q$ just means that there is a standard way to associate an element of $L^q$ to every linear bounded functional. this is done as already explained in the other answer to this post. I hope that now it is more clear :D

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    $\begingroup$ Very poor answer. What do you mean by "represented"? $\endgroup$ – Damian Sobota May 5 '13 at 22:00

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